高考数学复习第一篇求准提速基础小题不失分第16练圆锥曲线的定义方程与性质文市赛课公开课一等奖省名师优.pptx

第一篇求准提速,基础小题不失分,第,16,练圆锥曲线定义、方程与性质,1/62,明考情,圆锥曲线是高考热点，每年必考，小题中考查圆锥曲线定义、方程、离心率等，题目难度中等偏难,.,知考向,1.,圆锥曲线定义与标准方程,.,2.,圆锥曲线几何性质,.,3.,圆锥曲线综合,.,2/62,研透考点,关键考点突破练,栏目索引,明辨是非,易错易混专题练,演练模拟,高考押题冲刺练,3/62,研透考点,关键考点突破练,考点一圆锥曲线定义与标准方程,方法技巧,(1),椭圆和双曲线上点到两焦点距离能够相互转化，抛物线上点到焦点距离等于到准线距离,.,(2),求圆锥曲线方程惯用方法：定义法、待定系数法,.,4/62,A.8 B.10 C.12 D.15,1,2,3,4,5,答案,解析,5/62,1,2,3,4,5,6/62,1,2,3,4,5,答案,解析,7/62,且,P,(2,，,1),在渐近线上，,1,2,3,4,5,8/62,A.,x,2 B.,x,2,C.,x,1 D.,x,1,1,2,3,4,5,答案,解析,9/62,所以抛物线准线方程为,x,1,，故选,D.,1,2,3,4,5,10/62,1,2,3,4,5,答案,解析,11/62,由,AOF,是边长为,2,等边三角形得到,AOF,60,，,c,|,OF,|,2.,又,a,2,b,2,4,，,1,2,3,4,5,12/62,解析,由题意得焦点,F,(0,，,1),，,设,A,(1,，,3),，,则,|,MA,|,|,MF,|,|,MA,|,|,y,M,|,1,|,y,A,|,1,4.,4,1,2,3,4,5,答案,解析,13/62,考点二圆锥曲线几何性质,方法技巧,求离心率两种方法,(2),方程法：只需依据一个条件得到关于,a,，,b,，,c,各项式，然后两边同除以,a,或,a,2,得到关于,e,方程求,e,.,14/62,A.2 B.3 C.4 D.,与,取值相关,6,7,8,9,10,答案,解析,15/62,6,7,8,9,10,答案,解析,16/62,解析,依据题意，如图，设,F,(,c,，,0),，,6,7,8,9,10,a,2,b,2,c,2,，,17/62,6,7,8,9,10,5,a,5.,答案,解析,18/62,6,7,8,9,10,2,又,a,2,b,2,c,2,8,，,a,2.,答案,解析,解析,设,B,为双曲线右焦点，如图所表示,.,四边形,OABC,为正方形且边长为,2,，,19/62,10.,设抛物线,E,：,y,2,2,px,(,p,0),焦点为,F,，点,M,为抛物线,E,上一点，,|,MF,|,最小值为,3,，若点,P,为抛物线,E,上任意一点，,A,(4,，,1),，则,|,PA,|,|,PF,|,最小值为,_.,解析,由题意，,|,MF,|,最小值为,3,，得,3,，,p,6,，抛物线,E,：,y,2,12,x,，,抛物线,y,2,12,x,焦点,F,坐标是,(3,，,0).,设点,P,在准线上射影为,D,，,则依据抛物线定义可知,|,PF,|,|,PD,|,，,要求,|,PA,|,|,PF,|,取得最小值，即求,|,PA,|,|,PD,|,取得最小值，,当,D,，,P,，,A,三点共线时,|,PA,|,|,PD,|,最小，为,4,(,3),7.,7,6,7,8,9,10,答案,解析,20/62,考点三圆锥曲线综合,方法技巧,圆锥曲线范围，最值问题惯用方法,(1),定义性质转化法：利用圆锥曲线定义性质进行转化，依据平面几何中结论确定最值或范围,.,(2),目标函数法建立所求目标函数，将所求最值转化为函数最值处理,.,(3),条件不等式法：找出与变量相关全部限制条件，然后再经过处理不等式,(,组,),求变量范围,.,21/62,A.(,，,1),B.(,2,，,),答案,解析,11,12,13,14,15,16,22/62,假设焦点在,x,轴上，则,2,m,(,m,1),0,，,假设焦点在,y,轴上，则,(,m,1),2,m,0,，,11,12,13,14,15,16,23/62,12.(,四川,),设,O,为坐标原点，,P,是以,F,为焦点抛物线,y,2,2,px,(,p,0),上任意一点，,M,是线段,PF,上点，且,|,PM,|,2|,MF,|,，则直线,OM,斜率最大值为,答案,解析,11,12,13,14,15,16,24/62,显然，当,y,0,0,时，,k,OM,0,时，,k,OM,0.,要求,k,OM,最大值，不妨设,y,0,0,，,11,12,13,14,15,16,25/62,答案,解析,11,12,13,14,15,16,26/62,圆圆心为,(2,，,0),，半径为,2,，,解得,b,2,3,a,2,.,11,12,13,14,15,16,27/62,解析,11,12,13,14,15,16,答案,28/62,11,12,13,14,15,16,29/62,解析,由已知得直线方程为,y,2(,x,1).,11,12,13,14,15,16,答案,解析,30/62,16.,在直线,y,2,上任取一点,Q,，过,Q,作抛物线,x,2,4,y,切线，切点分别为,A,，,B,，则直线,AB,恒过定点,_.,(0,，,2),答案,解析,11,12,13,14,15,16,31/62,解析,设,Q,(,t,，,2),，,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,),，,又点,Q,(,t,，,2),坐标满足这两个方程，,11,12,13,14,15,16,32/62,所以直线,AB,恒过定点,(0,，,2).,11,12,13,14,15,16,33/62,1,2,3,4,明辨是非,易错易混专题练,解析,答案,34/62,1,2,3,4,35/62,2.,已知圆,C,1,：,(,x,3),2,y,2,1,和圆,C,2,：,(,x,3),2,y,2,9,，动圆,M,同时与圆,C,1,及圆,C,2,相外切，则动圆圆心,M,轨迹方程为,_.,1,2,3,4,答案,解析,36/62,解析,如图所表示，设动圆,M,与圆,C,1,及圆,C,2,分别外切于,A,和,B,.,依据两圆外切条件，,得,|,MC,1,|,|,AC,1,|,|,MA,|,，,|,MC,2,|,|,BC,2,|,|,MB,|,，,因为,|,MA,|,|,MB,|,，所以,|,MC,1,|,|,AC,1,|,|,MC,2,|,|,BC,2,|,，,即,|,MC,2,|,|,MC,1,|,|,BC,2,|,|,AC,1,|,2,6,，,所以点,M,到两定点,C,1,，,C,2,距离差是常数,.,又依据双曲线定义，,得动点,M,轨迹为双曲线左支,(,点,M,与,C,2,距离大，与,C,1,距离小,),，,其中,a,1,，,c,3,，则,b,2,8.,1,2,3,4,37/62,3.,若椭圆对称轴是坐标轴，且短轴一个端点与两个焦点组成一个正三,角形，焦点到同侧顶点距离为,，则椭圆方程为,_.,所以,b,2,a,2,c,2,9.,1,2,3,4,答案,解析,38/62,1,2,3,4,答案,解析,39/62,由椭圆几何性质，知a,c,|,PF,2,|0,，所以,m,3.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,42/62,A.2 B.6 C.8 D.14,解得,a,2,9,，,a,3,，,椭圆长轴长为,2,a,6,，,由椭圆定义可知，,|,PF,1,|,|,PF,2,|,6,，即,|,PF,2,|,2.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,43/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,44/62,1,2,3,4,5,6,7,8,9,10,11,12,45/62,1,2,3,4,5,6,7,8,9,10,11,12,由双曲线和圆对称性得四边形,ABCD,为矩形，,46/62,A.,m,n,且,e,1,e,2,1 B.,m,n,且,e,1,e,2,1,C.,m,n,且,e,1,e,2,1 D.,m,n,且,e,1,e,2,1,解析,由题意可得,m,2,1,n,2,1,，即,m,2,n,2,2,，,m,0,，,n,0,，故,m,n,.,e,1,e,2,1.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,47/62,5.,过抛物线,y,2,2,px,(,p,0),焦点作直线交抛物线于,P,，,Q,两点，若线段,PQ,中点横坐标为,3,，,|,PQ,|,10,，则抛物线方程是,A.,y,2,4,x,B.,y,2,2,x,C.,y,2,8,x,D.,y,2,6,x,解析,设抛物线,y,2,2,px,(,p,0),焦点为,F,，,P,(,x,1,，,y,1,),，,Q,(,x,2,，,y,2,),，,由抛物线定义可知，,线段,PQ,中点横坐标为,3,，,又,|,PQ,|,10,，,10,6,p,，可得,p,4,，,抛物线方程为,y,2,8,x,.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,48/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,49/62,即,a,2,b,2,7,，,联立,，解得,a,2,4,，,b,2,3,，,1,2,3,4,5,6,7,8,9,10,11,12,50/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,51/62,即,|,MF,2,|,3|,MF,1,|.,所以,b,2,a,2,，所以,c,2,b,2,a,2,2,a,2,，,1,2,3,4,5,6,7,8,9,10,11,12,52/62,又,B,，,D,，,M,三点共线，,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,53/62,1,2,3,4,5,6,7,8,9,10,11,12,所以,c,3,，得焦点为,F,1,(,3,，,0),，,F,2,(3,，,0).,依据椭圆定义，得,|,PM,|,|,PF,1,|,PM,|,(2,a,|,PF,2,|),10,(|,PM,|,|,PF,2,|).,因为,|,PM,|,|,PF,2,|,|,MF,2,|,，当且仅当,P,在,MF,2,延长线上时等号成立，,此时,|,PM,|,|,PF,1,|,最大值为,10,5,15.,15,解析,答案,54/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,(1,，,2),55/62,1,2,3,4,5,6,7,8,9,10,11,12,解析,设,P,(,x,，,y,),，由题设条件，,得动点,P,轨迹为,(,x,1)(,x,1),(,y,2)(,y,2),0,，,即,x,2,(,y,2),2,1,，它是以,(0,，,2),为圆心，,1,为半径圆,.,又,e,1,，故,1,e,2.,56/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,2,57/62,焦点,F,(0,，,1),，,抛物线,C,1,：,x,2,4,y,，准线方程为,y,1.,设点,M,在准线上射影为,D,，则依据抛物线定义可知,|,MF,|,|,MD,|,，,要求,|,MP,|,|,MF,|,取得最小值，,即求,|,MP,|,|,MD,|,取得最小值，,当,D,，,M,，,P,三点共线时，,|,MP,|,|,MD,|,最小，为,1,(,1),2.,1,2,3,4,5,6,7,8,9,10,11,12,58/62,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,59/62,1,2,3,4,5,6,7,8,9,10,11,12,解析,设,P,(,x,，,y,)(,y,0),，,60/62,1,2,3,4,5,6,7,8,9,10,11,12,整理得,(,x,c,),2,y,2,c,2,(,y,0),，,所以点,P,轨迹为以,(,c,，,0),为圆心，,c,为半径圆,(,去除两点,(0,，,0),，,(2,c,，,0),，,61/62,本课结束,62/62,